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Point (1, 0) is closest to which of the following lines? y =

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Point (1, 0) is closest to which of the following lines? y = [#permalink]

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New post 26 Aug 2008, 10:54
This topic is locked. If you want to discuss this question please re-post it in the respective forum.

Point (1, 0) is closest to which of the following lines?



y = x
y = 1
y + x = 3
x = 2
x + y = -1

Kudos [?]: 220 [0], given: 0

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Re: coordinate geometry [#permalink]

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New post 26 Aug 2008, 11:05
arjtryarjtry wrote:
Point (1, 0) is closest to which of the following lines?



y = x
y = 1
y + x = 3
x = 2
x + y = -1


Answer is A
A) y=x

Draw the lines in X-Y co-ordinate.

y = 1,x=2 lines are 1 unit away from (1, 0)
y + x = 3 , x + y = -1 are more than 1 unit away

draw perpendicular line which touchs the line y=x
distance is \(1/sqrt(2)\)
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Last edited by x2suresh on 26 Aug 2008, 11:18, edited 1 time in total.

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Re: coordinate geometry [#permalink]

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New post 26 Aug 2008, 11:05
arjtryarjtry wrote:
Point (1, 0) is closest to which of the following lines?



y = x
y = 1
y + x = 3
x = 2
x + y = -1


use the general formula. given line ax+by+c=0 and point (m,n), the perpendicular distance is
|am+bn+c|/sqrt(a^2 + b^2).

in this case, A seems to be the answer. Didnt calculate, so i could be wrong.

Kudos [?]: 68 [0], given: 0

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Re: coordinate geometry [#permalink]

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New post 26 Aug 2008, 11:07
I'd go for A.

Point 1,0 is 1/2sqrt2 from equation in A.

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Re: coordinate geometry [#permalink]

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New post 26 Aug 2008, 11:08
x2suresh wrote:
arjtryarjtry wrote:
Point (1, 0) is closest to which of the following lines?



y = x
y = 1
y + x = 3
x = 2
x + y = -1


Answer is A
A) y=x

Draw the lines in X-Y co-ordinate.

y = 1,x=2 lines are 1 unit away from (1, 0)
y + x = 3 , x + y = -1 are more than 1 unit away

draw perpendicular line which touchs the line y=x
distance is \(1/sqrt(3)\)


I think it should be 1/sqrt(2). coefficients of x and y would come into play. And both coefficients are 1.

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Re: coordinate geometry [#permalink]

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New post 26 Aug 2008, 11:17
bhushangiri wrote:
x2suresh wrote:
arjtryarjtry wrote:
Point (1, 0) is closest to which of the following lines?



y = x
y = 1
y + x = 3
x = 2
x + y = -1


Answer is A
A) y=x

Draw the lines in X-Y co-ordinate.

y = 1,x=2 lines are 1 unit away from (1, 0)
y + x = 3 , x + y = -1 are more than 1 unit away

draw perpendicular line which touchs the line y=x
distance is \(1/sqrt(3)\)


I think it should be 1/sqrt(2). coefficients of x and y would come into play. And both coefficients are 1.


you are right it should be \(1/sqrt(2)\) .. typo
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Re: coordinate geometry [#permalink]

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New post 12 Mar 2009, 15:17
bhushangiri wrote:
arjtryarjtry wrote:
Point (1, 0) is closest to which of the following lines?



y = x
y = 1
y + x = 3
x = 2
x + y = -1


use the general formula. given line ax+by+c=0 and point (m,n), the perpendicular distance is
|am+bn+c|/sqrt(a^2 + b^2).

in this case, A seems to be the answer. Didnt calculate, so i could be wrong.



Is there any other way to determine the perpendicular distance from a point to a line?

x2suresh,

How were you calculating the perpendicular distance from a point to a line? Were you using the same formula?

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Re: coordinate geometry [#permalink]

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New post 12 Mar 2009, 17:36
icandy wrote:
bhushangiri wrote:
arjtryarjtry wrote:
Point (1, 0) is closest to which of the following lines?



y = x
y = 1
y + x = 3
x = 2
x + y = -1


use the general formula. given line ax+by+c=0 and point (m,n), the perpendicular distance is
|am+bn+c|/sqrt(a^2 + b^2).

in this case, A seems to be the answer. Didnt calculate, so i could be wrong.



Is there any other way to determine the perpendicular distance from a point to a line?

x2suresh,

How were you calculating the perpendicular distance from a point to a line? Were you using the same formula?




I did it in different way.

slope of line y=x is 1
slop of line perpendicular to y=x is -1

y = -x +c -->perpendicular line .. which passes through (1,0)
--> c=1

y=-x +1 -->(2)
y=x -->(1)
find the intersection point (1) and (2) --> (1/2,1/2)

distance between (1/2,1/2) and (1,0) --> 1/sqrt(2)
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Re: coordinate geometry [#permalink]

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New post 14 Mar 2009, 01:06
The closest line i found is y=x. i.e A should be the answer.
Also the distance from y=x to the point (1,0) is 1/sqrt(2).

To find the distance from y=x, draw a line parallel to y=x which passes from (1,0).
Parallel line would be y=-x+c and as it passes from (1,0) we get c=1.
0=-1+c --> c=1
so the parallel line is y=-x+1.
as these two equations cross each other, we find out the point of intersection.
Using equations y=x and y=-x+1, we get (1/2,1/2) as the point of intersection.

now distance from point (1,0) to the point (1/2,1/2) = sqrt((1-1/2)^2+(0-1/2)^2) = 1/sqrt(2).

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Re: coordinate geometry [#permalink]

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New post 15 Mar 2009, 15:54
Here's maybe a silly question, but what i remember from back in school was the distance between two points was given by the formula here:

http://www.purplemath.com/modules/distform.htm

So for y=x, the point would be (1,1), and the distance from there to the point (1,0) would be 1 wouldnt it ?? Where have I misunderstood ?

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Re: coordinate geometry [#permalink]

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New post 15 Mar 2009, 23:11
I like bhushangiri's formula the best although for this question you dont really have to do any calculations. You can draw the axis and see approximate perpendicular distances of the point from the line.

Thanks.

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Re: coordinate geometry [#permalink]

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New post 16 Mar 2009, 07:01
if you plot it on the graph, it will be very clear that y=x is the closest to the point. and is 1/srrt 2 distance away perpendicularly.

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Re: coordinate geometry   [#permalink] 16 Mar 2009, 07:01
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Point (1, 0) is closest to which of the following lines? y =

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